Fn = P (1+i)n SPCAF = single-payment compound-amount factor

(1)

Module 2: Factors

SI-4251 Ekonomi Teknik

Muhamad Abduh, Ph.D.

Outline Modul 2



Engineering Economy Factors



Single-Payment Factors



Uniform-Series Present-Worth Factor



Capital-Recovery Factor



Sinking-Fund Factor



Gradient Formulas

Muhamad Abduh, Ph.D.

2-2 SI-4251 Ekonomi Teknik

Engineering Economy Factors



P

= present value, usually used to represent initial

investment



F

= future value, accumulated value at some future time



A =

a series of consecutive, equal, end-of-period

amount of payment



I

= interest, usually compounded



i

= interest rate per interest period, %



n

= number of (interest) periods



G

= arithmetic gradient, uniform arithmetic change in

magnitude of payment / receipts



g

= geometric gradient, constant rate of change in

magnitude of payment / receipts, %

Single-Payment Factors

0 1 2 3 n-1 n

P F

_?

F

_n

=

P

(1+

i

)

n SPCAF = single-payment compound-amount factor

From the opposite side, if F is given:

?

P

0 1 2 3 n-1 n

F

 

















_n

i

F

P

1

SPPWF = single-payment present worth factor Notation: (F/P, i, n)

(2)

Uniform-Series Present-Worth Factor

2 - 5 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.

?

P

SPPWF = single-payment present-worth factor

 

USPWF = uniform-series present-worth factor

Notation: (P/A, i, n)

Capital-Recovery Factor

2 - 6 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D. P

CRF = capital recovery factor

What A is needed to equal P at t=0?

Notation: (A/P, i, n)

?

Sinking-Fund Factor

0 1 2 3 n-1 n

SFF = sinking-fund factor

USCAF = uniform-series compound-amount factor

Notation: (A/F, i, n)

Notation: (F/A, i, n)

Standard Factor Notations

2 - 8

Factor Name Standard _Notation

Single-Payment Present-Worth SPPWF (P/F , i% , n)

(3)

Extreme Values

2 - 9

factor

_n

=

∞

;

i

known

n

=

0

;

i

known

( F / P, i % , n )

∞

1

( P / F, i % , n ) 0 1

( P / A , i % , n ) 1/i 0

( A / P, i % , n ) i NA

( A / F, i % , n ) 0 NA

( F / A , i % , n )

∞

0

Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik

Arithmetic Gradient (1)

2 - 10 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.  In some cases, periodic payments do not occur in an equal series, but they

maybe in a constant increase amount

 Equivalent with:

0 1 2 3 n-1 n

A1

A1 A1 A1 A1 A1 A1

3G 2G G

(n-3)G (n-2)G

(n-1)G

Arithmetic Gradient (2)

 Equivalent with:

0 1 2 3 n-1 n

A1

A1 A1 A1 A1 A1 A1

0 1 2 3 n-1 n

3G 2G G

(n-3)G (n-2)G

(n-1)G

+

Arithmetic Gradient (3)

 Equivalent with: A = A1 + A2

0 1 2 3 n-1 n

A1

A1 A1 A1 A1 A1 A1

+

0 1 2 3 n-1 n

A2

A2 A2 A2 A2 A2 A2

=

0 1 2 3 n-1 n

(4)

Arithmetic Gradient (4)

2 - 13 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.  Where:

A1 = payment at the end of the first year

G = gradient, annual change

n = number of period

A = equivalent equal annual payment A = A1 + A2

 

  

 

   

1 1 1

2 n

i n i G A

UGSF = uniform-gradient-series factor

Arithmetic Gradient Formulas

 

  

  

   

 

  

 

 n n

n

i n i i

i i G P

1 1

1 1 1

UGPWF = uniform-gradient-present-worth factor

 

    

  

   

 

  

 n

i i i G F

n ₁

1 1

UGFWF = uniform-gradient-future-worth factor

Negative Arithmetic Gradient (1)

2 - 15 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.  In some cases, periodic payments do not occur in an equal series, but they

maybe in a constant decrease amount

0 1 2 3 n-1 n

Arithmetic Gradient (2)

2 - 16 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.  Equivalent with:

0 1 2 3 n-1 n

A1

0 1 2 3 n-1 n

3G 2G G

(n-3)G (n-2)G

(n-1)G

-

(5)

Arithmetic Gradient (3)

2 - 17 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.  Equivalent with: A = A1– A2

Geometric Gradient

2 - 18 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D. 

In many situations, periodic payments decreases or increases

not by a constant amount, but at a constant percentage











GGSF = geometric-gradient-series factor F1(1+g)n-1

growth-free rate

Using Factors

Finding Given Factor Equation Formula

P F ( P / F , i % , n ) P = F (P/F, i%, n)

Interest Table

 For the shake of simplicity and easier calculation all values based on interest formulas have been tabularized

 For any values that are not available in the interest table, interpolation or extrapolation can be applied:

n Single Payment Equal Payment Series

Uniform Gradient Series

(F/P, i, n) (P/F, i, n) (F/A, i, n) (A/F, i n) (P/A, i n) (A/P, i, n) (A/G, i n)

12% interest factors for continuous compounding

1.352

(6)

Spreadsheet Formulas

Factors in Engineering Economy

Excel Functions

P = Present Value = PV (i%,n,A,F)

F = Future Value = FV (i%,n,A,P)

A = Equal, Periodic Value = PMT (i%,n,P,F)

n = Number of Periods = NPER (i%,A,P,F)

i = Compound Interest Rate = RATE (n,A,P,F)

Exercises (1)

22

1.

How much money should you be willing to pay now for a

guaranteed $600 per year for 9 years starting next year, at a

rate of return of 16% per year?

2.

Determine the value of the A/P factor for an interest rate of

7.3% and n of 10 years, that is, (A/P, 7.3%, 10).

3.

HP has completed a study indicating that $50,000 in reduced

maintenance this year on one processing line resulted from

improved integrated circuit (IC) fabrication technology

based on rapidly changing designs.

a. If HP considers these types of savings worth 20% per year, find the

equivalent value of this result after 5 years.

b. If the $50,000 maintenance savings occur now, find its equivalent

value 3 years earlier with interest at 20% per year.

Exercises (2)

23

4.

Three contiguous counties in Florida have agreed to pool tax

resources already designated for county-maintained bridge

refurbishment. At a recent meeting, the county engineers

estimated that a total $500,000 will be deposited at the end

of the next year into an account for the repair of old and

safety-questionable bridges throughout the three-county

area. Further, they estimate that the deposits will increase by

$100,000 per year for only 9 years thereafter, then cease.

Determine the equivalent:

a. Present worth if county funds earn interest at a rate of 5% per year.

b. Annual series amounts if county funds earn interest at a rate of 5%

per year.

Exercises (3)

24

5. How long will it take for $1000 to double if the interest rate is 5% per year?

6. Engineers at SeaWorld, a division of Busch Garden, Inc., have completed an innovation on an existing water sports ride to make it more exciting. The modification costs only $8,000 and is expected to last 6 years with a $1,300 salvage value for the solenoid mechanisms. The maintenance cost is expected to be high at $1,700 the first year, increasing by 11% per year thereafter. Determine the equivalent present worth of the modification and maintenance cost. The interest rate is 8% per year.

7. If Laurel can make an investment in a friend’s business of $3,000 now in

(7)

Exercises (4)

25

8. An engineering technology group just purchased new CAD software for $5,000 now and

annual payment of $500 per year for 6 years starting 3 years from now for annual upgrades. What is the present worth of the payments if the interest rate is 8% per year?

9. Recalibration of sensitive measuring devices costs $8,000 per year. If the machine will be recalibrated for each of 6 years starting 3 years after purchase, calculate the 8-year equivalent uniform series at 16% per year.

10. An engineering company in Wyoming that owns 50 hectares of valuable land has decided to lease the mineral rights to a mining company. The primary objective is to obtain long term income to finance ongoing projects 6 and 16 years from the present time. The engineering company makes a proposal to the mining company that it pay $20,000 per year for 20 years beginning 3 years from now, plus $10,000 six years from now and $15,000 sixteen years from now. Determine the equivalent values listed below at 16% per year:

1. Total present worth in year 0.

2. Future worth in year 22.

3. Annual series over all 22 years.

4. Annual series over the first 10 years.

5. Annual series over the last 12 years.

Exercises (5)

26

11. Gerri, an engineer at Fujitsu, Inc., has tracked the average inspection cost on a robotics manufacturing line for 8 years. Cost averages were steady at $100 per completed unit for the first 4 years, but have increased consistently by $50 per unit for each of the last 4 years. Gerri plans to analyze the gradient increase using the P/G factor. Where is the present worth located for gradient? What is the general relation used to calculate total present worth in year 0?

12. Chemical engineers at a Coleman Industries plants in the Midwest have determined that a small amount of a newly available chemical additive

will increase the water repellency of Coleman’s tent fabric by 20%. The

plant superintendent has arranged to purchase the additive through a 5-year contract at $7,000 per 5-year, starting 1 5-year from now. He expects the annual price to increase by 12% per year thereafter for the next 8 years. Additionally, an initial investment of $35,000 was made now to prepare a site suitable for the contractor to deliver the additive. Use i = 15% to determine the equivalent total present worth for all cash flows.

Exercises (6)

13. Assume that you are planning to invest money at 7% per year as shown by the increasing gradient of the following figure. Further, you expect to withdraw according to the decreasing gradient shown. Find the net present worth and equivalent annual series for the entire cash flow sequence and interpret the results.

0 1 2 3 4 5 6 7 8 9 10 11 12 $2000

$2500 $3000

$3500 $4000

$5000 $4000

$3000 $2000

$1000 $1000 $1000

Homework #2

